- Given a collection of intervals, find the minimum number of intervals you need to remove to make the REST of the intervals non-overlapping.
- Note:
- You may assume the interval's end point is always bigger than its start point.
- Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
- Example 1:
- Input: [ [1,2], [2,3], [3,4], [1,3] ]
- Output: 1
- Explanation: [1,3] can be removed and the REST of intervals are non-overlapping.
- Example 2:
- Input: [ [1,2], [1,2], [1,2] ]
- Output: 2
- Explanation: You need to remove two [1,2] to make the REST of intervals non-overlapping.
- Example 3:
- Input: [ [1,2], [2,3] ]
- Output: 0
- Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
求移除多少区间后, 剩余区间都是不重叠的.
先求最多能组成多少不重叠的区间, 再用总区间数减去不重叠区间数.
C++:
- /**
- * Definition for an interval.
- * struct Interval {
- * int start;
- * int end;
- * Interval() : start(0), end(0) {}
- * Interval(int s, int e) : start(s), end(e) {}
- * };
- */
- bool compare(const Interval& a ,const Interval& b){
- return a.end <b.end ;
- }
- class Solution {
- public:
- int eraseOverlapIntervals(vector<Interval>& intervals) {
- if (intervals.size() == 0){
- return 0 ;
- }
- sort(intervals.begin() , intervals.end() , compare) ;
- int cnt = 1;
- int end = intervals[0].end ;
- for(int i = 1 ; i < intervals.size() ; i++){
- if (intervals[i].start < end){
- continue ;
- }
- end = intervals[i].end ;
- cnt++ ;
- }
- return intervals.size() - cnt ;
- }
- };
来源: http://www.bubuko.com/infodetail-2935737.html